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3 years ago

15

How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 36.0°C greater th

an when they were laid? Their original length is 26.0 m.

1 answer:

JulijaS [17]3 years ago

7 0

Answer:

11 mm

Explanation:

Original length L' = 26 m

Increase in temperature dT = 36.0°C

The gap l =?

The coefficient of linear expansion for steel & = 12 × 10^−6 per °C

The gap will be gotten from the equation of linear expansion.

l = L' x & x dT

substituting, we have

L = 26 x 12 × 10^−6 x 36

L = 0.011 m = 11 mm

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A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

NikAS [45]

Answer:

The gazelles top speed is 27.3 m/s.

Explanation:

Given that,

Acceleration = 4.2 m/s²

Time = 6.5 s

Suppose we need to find the gazelles top speed

The speed is equal to the product of acceleration and time.

We need to calculate the gazelles top speed

Using formula of speed

$v=at$

Where, v = speed

a = acceleration

t = time

Put the value into the formula

$v=4.2\times6.5$

$v=27.3\ m/s$

Hence, The gazelles top speed is 27.3 m/s.

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2 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

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3 years ago

A sewing machine uses 1.6kWh of electricity in one day. If electricity costs 9p per unit, what is the total cost in pence of usi

atroni [7]

Answer:

$\huge\boxed{\sf 1.6\ units = 14.4p}$

Explanation:

Since,

<h3><u>1 kWh = 1 unit</u></h3>

So,

1.6 kWh = 1.6 units

If,

<h3>1 unit = 9p</h3>

1.6 units = 9p × 1.6

1.6 units = 14.4p

$\rule[225]{225}{2}$

8 0

1 year ago

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0

3 years ago

What does AC stand for in physics?

Svetradugi [14.3K]

Answer:

Alternating current

3 0

2 years ago

Read 2 more answers